A 2.00 kg pendulum bob on a string 1.50 m long, is released with a velocity of 4.00 m/s when the support string makes an angle of 30.0 degrees with the vertical. What is the speed of the bob at the bottom of the swing?
A) 4.47 m/s
B) 4.02 m/s
C) 3.75 m/s
D) 3.50 m/s
E) 2.85 m/s

Question

Quizplus · Accepted Answer

The speed of the bob at the bottom of the swing can be found using the principle of conservation of mechanical energy. The initial mechanical energy of the system is the sum of the gravitational potential energy (when the pendulum is released) and the kinetic energy due to its velocity at that point. At the bottom of the swing, all this energy is converted into kinetic energy.The initial gravitational potential energy (U) can be calculated using the height (h) change as the pendulum swings down. Since the pendulum makes a 30-degree angle with the vertical, the height change can be found using trigonometry (h = L - Lcos(θ)), where L is the length of the string and θ is the angle with the vertical. The initial kinetic energy (K) is given by 1/2 m v^2, where m is the mass and v is the velocity.At the bottom of the swing, the potential energy is 0 (since we consider this point as the reference height), and all the energy is kinetic. By equating the initial mechanical energy to the final kinetic energy, we can solve for the final speed.Given:- Mass (m) = 2.00 kg- Length (L) = 1.50 m- Initial velocity (v) = 4.00 m/s- g = 9.81 m/s^2 (acceleration due to gravity)- θ = 30 degreesInitial height change (h) = L - Lcos(θ) = 1.50 - 1.50cos(30) = 1.50 - 1.50(√3/2) ≈ 0.25 mInitial potential energy (U) = mgh = 2.00 * 9.81 * 0.25 ≈ 4.905 JInitial kinetic energy (K) = 1/2 m v^2 = 0.5 * 2.00 * (4.00)^2 = 16 JTotal initial mechanical energy = U + K ≈ 4.905 J + 16 J = 20.905 JAt the bottom, all this energy is kinetic, so:1/2 m v_f^2 = 20.905 JSolving for v_f gives:v_f = sqrt((2 * 20.905) / 2.00) ≈ 4.47 m/sTherefore, the speed of the bob at the bottom of the swing is approximately 4.47 m/s.

A 200 Kg Pendulum Bob on a String 1